Section 14.7 Completeness
Given a vector and an orthonormal basis, it is easy to determine the components of the vector in the given basis. For example, if
\begin{equation}
\FF = F_x \,\xhat + F_y \,\yhat + F_z \,\zhat\tag{14.7.1}
\end{equation}
then dotting both sides of this equation with \(\xhat\) yields the \(x\)-component of the vector, namely \(F_x\text{.}\)
\begin{align}
\xhat\cdot\FF
\amp= \xhat\cdot(F_x \,\xhat + F_y \,\yhat + F_z \,\zhat)\notag\\
\amp= F_x\, \cancelto{1}{\xhat\cdot\xhat}
+ F_y \cancelto{0}{\xhat\cdot\yhat}
+ F_z \cancelto{0}{\xhat\cdot\zhat}\notag\\
\amp= F_x\tag{14.7.2}
\end{align}
where the orthonormality of the basis vectors has been used in the middle step.
Put differently,
\begin{equation}
\FF = (\FF\cdot\xhat)\,\xhat + (\FF\cdot\yhat)\,\yhat
+ (\FF\cdot\zhat)\,\zhat\text{.}\tag{14.7.3}
\end{equation}
All we need to make this idea work is an orthonormal basis. (If the basis is not normalized, the method still works, but we will need to account for the normalization of the basis vectors explicitly.)
In function spaces, for example for Fourier series in
Section 14.6, the sum in the dot product above becomes an integral. The orthogonal basis elements are the trigonometric functions
\begin{equation}
\left\{1,\cos nx,
\sin nx\right\}\tag{14.7.4}
\end{equation}
with \(n\) a positive integer. There are an infinite number of basis vectors! But what vector space do they span?
Definition 14.8. Completeness.
For a finite dimensional vector space, it is usually obvious how many vectors are required to make a basis, i.e. the minimum number of vectors you need to ensure that every vector can be built out of a linear combiniation of the basis. For example, for arrows in 3-dimensional space, we need three basis vectors. That’s what the number 3 in “3-dimensional” means. The technical language for this concept is that this basis is
complete. For an infinite dimensional vector space, such as periodic functions, the proof of what makes a basis complete is more difficult. Fortunately, mathematicians have all ready done the proofs in the applied cases that we care about, so it is only necessary to know the result. It turns out that the any periodic function that is reasonably well behaved (for instance, piecewise smooth) can be expanded uniquely in terms of the basis given in
(14.7.4). This basis is complete on the vector space of (piecewise smooth) periodic functions with period
\(2\pi\text{.}\)
